compare two matrices or vectors that have the same numbers but in different order

I wish to comapre two vectors or matrices and see if they have the same numbers but they may not be in the same order such as:

1 2 3 and 2 3 1

4 5 6 4 6 5

7 8 9 8 9 7

or

1 2 3 4 5

vs

3 4 2 5 1

Answer 1

[Star Strider]

The easiest way is to sort them across rows:

A = [1 2 3
    4 5 6
    7 8 9];

B = [2 3 1
    4 6 5
    8 9 7];

L = all(sort(A,2) == sort(B,2),2)

producing:

L =

  3×1 logical array

   1
   1
   1

Note that you get a much different result if you do not sort them.

[Andrew Luce]

Now let me add the caveat of saying what if you want to compare two matrices to see if their numbers are within a certain tolerance ...In the case given below +-0.5 in addition to not being in order 1 2 3 2.3 2.8 1.2 4 5 6 5.4 5.7 3.9

[Star Strider]

Without a specific example (only one vector here, so considering two matrices with elements in each row that are close though not exactly equal), I would sort them and then use the ismembertol (link) function, for example:

L = ismembertol(sort(A,2),sort(B,2), 0.6, 'ByRows',1)

Choose whatever tolerance is appropriate (I use 0.6 here), allowing some additional tolerance to complensate for floating-point approximation error.

[Andrew Luce]

Hello

Last caveat, what if the values were not in any order, where the values that are roughly 1 2 3 are not all contained on the first row

A=[1 2 3 B=[8.9 1.9 4.6

4 5 6 1.2 3.9 2.8

7 8 9] 8.1 7.3 5.7

Thank you for all your help

Andrew

[Star Strider]

As always, my pleasure.

This becomes a bit more complicated, and I’m not certain what result you want here. I would continue to use ismembertol, and set the tolerance depending on what you want to consider.

For example:

A=[1 2 3
   4 5 6
   7 8 9];

B=[8.9  1.9  4.6
   1.2  3.9  2.8
   8.1  7.3  5.7];

L = ismembertol(A, B, 0.025) 

produces:

L =

  3×3 logical array

   1   1   1
   1   0   0
   0   1   1

This is an ‘any’ condition, where any elements of ‘A’ are within tolerance of the elements in ‘B’. You could then change the tolerance or do logical comparisons on the outputs.

Note that you can also use ismembertol with a second output that would provide more information:

   [L,Locb] = ismembertol(A, B, 0.025)

producing:

L =
   1   1   1
   1   0   0
   0   1   1

Locb =
     2     4     8
     5     0     0
     0     3     1

The ‘Locb’ output contains the index location in ‘B’ for each element in ‘A’ that is a member of ‘B’.

In this instance, sorting the rows (or columns) would not provide any benefit.